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ABC is a triangle inscribed in a circle, AC being the diameter of the circle. The length of AC is as much more than the length of BC as the length of BC is more than the length of AB. Find AC:AB.


ABC is a triangle inscribed in a circle, AC being the diameter of the circle. The length of AC is as much more than the length of BC as the length of BC is more than the length of AB. Find AC:AB.

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3 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
11 years ago

Dear Vipul,

Solution:- Given, AC = a BC and BC = a AB. We want to find AC: AB  = a2

Since, AC is the diameter of the circle, it subtends 900 angle at B. Let, AC = 'd' and angle C be 'x'. Then,

AB = d sinx , and BC = d cosx

So, AC = a BC will give 1 = a cosx     or    cosx = 1/a

And, BC = a AB will give cos x = a sinx

Since, cos x = 1/a, sin x = √(a2 -1) / a

So, 1 = a √(a2 -1)          or         a4 - a2 -1 = 0

On solving quad. eq., we'll get - a2 = (√5 +1) / 2  [ANS]

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Regards,
Askiitians Experts
Priyansh Bajaj

vipul agrawal
19 Points
11 years ago

Dear Priyansh,

I would like to correct your solution as follows:-

As per question we have to assume that if AB = X and BC = X+k then AC = BC+ k or AC = X+2k. Then we have to find AC:BC which is equal to 5:3. I would request you to find it out for my son Vipul.

Thanking you.

G.P.Agrawal

F/o Master Vipul Agrawal.

Siddhant Chakraborty
11 Points
3 years ago
ΔABC is an right angle triangle as Ac is the diameter of the circle. 
Now it is given that AC-BC=BC-AB
AC+AB=2BC
AC=2BC-AB............(1)
AB²+BC²=AC². .........(Pythagoras theorem) 
AB²+BC²=(2BC-AB)²............(using eqn.  1)
AB²+BC²=4BC²+AB²-4*BC*AB
BC²-4BC²=-4*BC*AB
3BC²=4*BC*AB
3BC²/4BC=AB
3/4BC=AB........(2)
AB²+BC²=AC²
(3/4BC)²+BC²=AC².......(Using eqn. 2)
9/16BC²+BC²=AC²
9/16BC²+16/16BC²=AC²
25/16BC²=AC²
5/4BC=AC........(3)
AC:AB=AC/AB
(5/4BC)/(3/4BC)..........(Using eqn.  2 &3)
5/3=AC/AB=AC:AB

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